 ಎಲಾ ಎಲಿಭಸ ಆಲಿಹತಯತಿ ಜರಕಿ ಮನಕಿಮಾ ಎಲಿ ಹೆಲ೅ನಆ� returning to thevil👀ಂದ� prototy obstinate pletsami ನಿಯಯಚಿ Ãಮೆ cant and ?! ಣನಲನಂ organizes the polynomial equation is A ఆరమాలెయాల్ పంనితిటణి . సింని వారోనిల్ ఩కారాని టపన్కారానుని. పచడింని వారోని పంలాపలరటివారాన్లా. f of x is a polynomial of degree greater than equal to 1 whose coefficients are real and complex number then f of x equal to 0 is said to be a polynomial equation polynomial equation of degree 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 where a0 not equal to 0 similarly polynomial equation of degree n is a0 x0 n plus a1 x0 n minus 1 plus a2 x0 n minus 2 plus dot dot plus a n minus 1 x plus a n equal to 0 where a0 not equal to 0 root of a polynomial equation if alpha is a value of x such that f of alpha is 0 then alpha is said to be root of a equation f of x equal to 0 factor theorem so factor theorem term in a k define coro if alpha is a root of a polynomial equation f of x equal to 0 then polynomial equation f of x is divisible by x minus alpha and conversely theorem 2 if every equation of n degree has n roots and no more these 2 are important term related to factor theorem relation between roots and coefficient zodium f of x at a polynomial of degree n consider coro so f of f of x equal to a0 x0 n plus a1 x0 n minus 1 plus dot dot a n this is a polynomial of degree n and let alpha 1 alpha 2 alpha 3 dot dot alpha n be the root of the equation f of x equal to 0 where summation alpha 1 summation alpha 1 is equal to sum of the roots is equal to a1 by a0 summation alpha 1 alpha 2 sum of the product of the roots taken 2 at a time is equal to minus a2 by a0 summation alpha 1 alpha 2 alpha 3 is equal to sum of product of roots taken 3 at a time is equal to a3 by a0 thick then again me product of n roots alpha 1 alpha 2 alpha 3 dot dot alpha n is equal to minus 1 to the power n a n divided by a0 so this is the relation between roots and coefficients now we discussed some particular case relation between roots and coefficient of quadratic cubic and bio quadratic equation are deduced as follows for quantity equation a0 x square plus a1 x plus a2 is equal to 0 where a0 not equal to 0 so here summation alpha 1 in sum of the roots is equal to alpha plus beta is equal to minus a1 by a0 product of roots alpha 1 alpha 2 is equal to alpha beta is equal to a2 by a0 similarly cubic equation a0 x2 plus a1 x square plus a2 x plus a3 is equal to 0 a0 not equal to 0 here sum of the roots alpha plus beta plus gamma is equal to minus a1 by a0 summation alpha 1 alpha 2 is equal to alpha beta beta gamma alpha is equal to a2 by a0 alpha 1 alpha 2 alpha 3 is equal to alpha beta gamma is equal to minus a3 by a0 similarly for bio quadratic equation so relation between roots and coefficients are here we consider bio quadratic equation a0 x2 over 4 plus a1 x2 plus a2 x2 plus a3 x plus a4 is equal to 0 a0 not equal to 0 here sum of roots alpha plus beta plus gamma plus delta is equal to minus a1 by a0 similarly alpha beta alpha delta beta gamma beta delta plus alpha delta is equal to a2 by a0 and sum of product of roots taken 3 at a time means alpha beta gamma plus beta gamma alpha plus gamma delta alpha plus gamma beta delta is equal to minus a3 by a0 and product of all roots alpha beta gamma delta is equal to a4 by a0 so these are the three cases which we deduce the relation between roots and coefficients first one is quadratic second one cubic and third is bio quadratic so now we discuss example related to the concept relation between roots and coefficients solved the equation x cube minus 7 x2 minus 5x plus 75 is equal to 0 2 of each roots being equal now we will give an example of this equation solved the equation x cube minus 7 x2 minus 5x plus 75 is equal to 0 2 of each roots being equal so I mean at a this is a 2 of each roots are equal so let the roots of the equation be alpha comma alpha beta because 2 roots are equal so we consider 2 roots alpha alpha equal so now summation alpha is equal to alpha plus alpha plus beta is equal to twice alpha plus beta equal to 7 so here from this equation the relation between roots are alpha plus beta plus gamma equal to minus a1 by a0 so here we get alpha plus alpha plus beta this is equal to twice alpha plus beta is equal to 7 this implies beta equal to 7 minus twice alpha next we consider second relation alpha plus beta gamma plus gamma alpha is equal to a2 by a0 so here we get alpha into alpha this alpha square alpha beta beta alpha is equal to minus 5 so here we get alpha square twice alpha so alpha square plus twice alpha beta is equal to minus 5 so alpha square equal to twice alpha 7 minus twice alpha from here beta is equal to 7 minus twice alpha so we put here beta 7 minus twice alpha so we get here twice alpha break it 7 minus twice alpha plus 5 is equal to 0. So, simplifying this relation we get minus 3 alpha square plus 14 alpha plus 5 is equal to 0. So, multiplying by minus 1 we get 3 alpha square minus 14 alpha minus 5 is equal to 0. So, now we factorize this quality equation. So, we get 3 alpha square minus 15 alpha plus alpha minus 5. So, from here we get factorizing this we get alpha minus 5 3 alpha plus 1 is equal to 0. So, from here we get 2 value of alpha. So, alpha is equal to 5 minus 1 by 3. So, when alpha first we consider alpha equal to 5. So, when alpha is equal to 5 beta is equal to 7 minus 2 into 5 this means 7 minus 10 minus 3. When alpha is equal to minus 1 by 3 we get beta 7 minus 2 into minus 1 by 3. So, 3 by 3. So, when alpha is equal to 5 we get root char from here 5 5 and beta we get minus 3. So, 5 comma 5 minus 3 and when we consider alpha is equal to minus 3. So, we get root char minus 1 by 3 comma minus 1 by 3 23 by 3. So, again we have another relation potak of roots. So, alpha beta gamma is equal to minus a 3 by a 0. So, here we get potak of the roots alpha beta gamma is equal to alpha into alpha into beta is equal to alpha is equal to beta is equal to minus 75. Since, 5 into 5 into 5 into minus 3 if we multiply this 3 roots. So, is equal to 75. So, therefore, this 3 roots satisfies this relation, but if we put alpha is equal to minus 1 by 3 alpha is equal to minus 1 by 3 beta is equal to what 23 by 3 then we get this potak of this, but this not is equal to minus 75. So, required root char 5 comma 5 comma minus 3. So, this is the required root char. So, now define symmetric function of roots. A function of the roots of a equation is said to be symmetric if it remains unaltered by an intersense of any 2 of its roots. For example, if alpha beta gamma with the roots of a cubic equation a naught x cube x cube plus a 1 x square plus a 2 x plus a 3 equal to 0 a naught not equal to 0 then here we get summation alpha is equal to alpha plus beta plus gamma summation alpha square is equal to alpha square plus beta square plus gamma square summation alpha beta is equal to alpha beta beta gamma plus gamma alpha. So, these are the example of symmetric function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . so we have here we have to find summation alpha square plus alpha square beta number two summation alpha square beta square and summation alpha by beta plus beta by alpha so here we consider three roots alpha beta gamma these roots so we know that summation alpha means alpha plus beta is equal to p summation beta is equal to alpha beta plus beta gamma plus gamma alpha is equal to q and alpha beta gamma is equal to minus i so now firstly calculate summation alpha into summation alpha beta their summation alpha is equal to p submission alpha beta is equal to q now if you if we multiply this x2 expression we get alpha plus beta plus gamma, alpha beta beta gamma plus gamma alpha is equal to pq. Multiplying these two expressions we get summation alpha square beta plus 3 alpha beta gamma is equal to pq. Here we get summation alpha square beta and alpha beta gamma is equal to r, so here we get 3r. Next summation alpha square beta is equal to pq minus r, so therefore value of summation alpha square beta is equal to pq minus 3r. Next we have to find summation alpha square beta square. Summation alpha square beta square means alpha square beta square plus beta square gamma square plus gamma square alpha square. Now we get a plus b plus c is equal to a square plus b square plus c square to a b plus b c plus c. So here we get a square plus b square plus c square is equal to a plus b plus c square minus 2 a b plus b c plus c. In formula we have to apply when we pump alpha beta beta gamma plus gamma alpha holy square is equal to minus 2 alpha beta square gamma plus beta gamma square alpha plus gamma beta alpha square. So here I pump q square q, so q square here alpha beta gamma I commonly pump 2 alpha beta gamma bracket alpha plus beta plus gamma. So value of alpha beta gamma is r, so 2r and alpha beta gamma is p, so q square minus 2pr. Next we will get summation alpha by beta plus beta by alpha. So summation alpha by beta plus beta by gamma I say that I break expression to break point I pump alpha by beta plus beta by alpha plus bracket beta by gamma plus gamma by beta plus bracket alpha by alpha plus alpha by gamma. So here for me alpha beta gamma I say I pump alpha square gamma plus beta square gamma plus alpha beta square plus alpha gamma square plus beta gamma square plus alpha square beta. So this expression for me alpha beta square alpha square beta I write the expression and I will get alpha beta common long common only alpha beta bracket alpha plus beta. So alpha beta bracket alpha plus beta gamma plus beta gamma plus alpha beta divided by r. So this expression for me alpha beta bracket alpha plus beta gamma plus gamma alpha minus 3 alpha beta. So p here from q pump alpha beta is r, so p p minus 3r. So this is the value of summation alpha bracket alpha beta plus beta by gamma. So here for concept of me here alasana krim, a concept of our synthetic division. To find question and remember when a given polynomial is divided by binomial, binomial easily we use synthetic division. So synthetic division that I am example alasana krim solve the cubic equation x cube minus 5x square minus 4x plus 20 is equal to 0. Given that product of 2 roots is 10. So synthetic division of higher I mean a cubic equation to homonyl krim. So I mean example 3 alasana krim. Let alpha beta gamma be the root of the equation. We know that alpha beta gamma is equal to minus 20 and alpha beta is equal to 10 and this is given. So therefore simplifying this equation to this relation we get gamma equal to minus 2. Putting the value of alpha beta here we get gamma equal to minus 2. So minus 2 is the one root of this equation given equation. Hence x plus 2 is a factor of x cube minus 5x square minus 4x plus 20. So now we have to find other 2 roots. So by synthetic division applying synthetic division. So here x plus 2 is a factor one factor so minus 2. So here coefficient of x cube is one coefficient of x square minus 5 coefficient of x minus 4 and constant term 20. So this one we take here one this one. So now minus 2 into 1 is equal to minus 2 and minus 5 plus minus 2 is equal to minus 7. So again minus 2 into minus 7 is equal to 14. Then minus 4 plus 14 10. So again minus 2 into 10 so minus 20. So 20 plus minus 20 is equal to 0. So this is the remainder. So remainder is equal to 0. So we get equation of this equation is equal to x square minus 7x plus 10. So this is the quantity equation. So factorizing this quantity equation we get x square minus 5x minus 2x plus 10. So factorizing we get x minus 5 into x minus 2. So now x square minus 7x plus 10 is equal to 0. So x minus 5 into x minus 2 is equal to 0. So x equal to 5 and x equal to 2. so other 2 roots are 2.5 so therefore we get 3 roots is equal to 2, 5 and minus 4 today in our video we discussed the theory of equation 1 so we discussed the theory of equation 1, factor theorem of a polynomial equation, today we discussed the theory of equation 1, relation between roots and coefficient of a polynomial equation. now we discussed the synthetic division for a polynomial equation. so we will end today, Namaskar